Proof

By hypothesis, $(X,\mathcal O)$ is a topological space and $B\subset X$ its subset.

Let $x\in B \setminus \delta B$ be an element of the interior.
Since $x$ is not a boundary point, there is a neighborhood $N_x$ that has no points in common with the complement $B^C$.
Moreover, we can choose $N_x$ in a way that it has even no points in common with boundary $\delta B.$ Otherwise, if there existed an $y\in N_x \cap \delta B,$ then $N_x$ would be a neighborhood of $y$ and would have no points in common with the complement $B^C.$ But every neighborhood of a boundary point has points in common with the complement.
Thus, $N_x$ has no boundary point in it.
Altogether, we have shown that every neighborhood of $x$ is contained in the interior.
It follows that the interior is open.

We have $\delta B=(B\cup \delta B)\setminus(B\setminus \delta B).$
It follows that $X\setminus \delta B=(X\setminus(B\cup\delta B))\cup(B\setminus \delta B),$ which is, according to $ (1)$ and $(2)$, a union of two open sets.
By the third axiom of topology, $X\setminus \delta B=(\delta B)^C$ is open.
Therefore, the complement $((\delta B)^C)^C)=\delta B$ is closed, which is the boundary.
∎

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Forster Otto: "Analysis 2, Differentialrechnung im $\mathbb R^n$, Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984